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How is $p\ '(x)$ written in that solution? I understand that $p\ '(x)$ will be some $4$ degree polynomial but how did we find out that $2$ and $-2$ will be roots of $p\ '(x)$?

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    This is explicitly **not** an algebra/precalculus problem. – 2017-02-23
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    Will anyone care to tell why this question has been downvoted? – 2017-02-23
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    Probably because it was miscategorized. – 2017-02-23

2 Answers 2

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By the Division Algorithm, we know that there are some degree $2$ polynomials $Q_1, Q_2$ such that: $$ P(x) = Q_1(x) \cdot (x - 2)^3 - 2 = Q_2(x) \cdot (x + 2)^3 + 2 $$

Differentiating each equation with product rule, we get: \begin{align*} P'(x) &= Q_1'(x) \cdot (x - 2)^3 + Q_1(x) \cdot 3(x - 2)^2 = (x - 2)^2 \cdot (\underbrace{(x - 2) Q_1'(x) + 3Q_1(x)}_{\textsf{some polynomial of degree $2$}}) \\ \\ P'(x) &= Q_2'(x) \cdot (x + 2)^3 + Q_2(x) \cdot 3(x + 2)^2 = (x + 2)^2 \cdot (\underbrace{(x + 2) Q_2'(x) + 3Q_2(x)}_{\textsf{some polynomial of degree $2$}}) \end{align*}

But then since $(x - 2)^2$ and $(x + 2)^2$ are coprime factors of $P'(x)$ and since $\deg P' = 4$, it follows that: $$ P'(x) = K(x - 2)^2(x + 2)^2 $$ for some constant $K$, as desired.

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    How are they coprime factors? Even if they are coprime factors how does it imply the next step – 2017-02-23
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    @Cotton I believe by coprime, he means there is no polynomial that divides both $(x-2)^2$ and $(x+2)^2$. He's shown $P'(x)$ has $2$ roots of $2$ and $2$ roots of $-2$. That accounts for all $4$ roots. – 2017-02-23
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    @Mike How do we know that p(x) divides (x-2) and (x+2) in the first place? – 2017-02-23
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    @Cotton Do you mean how do we know that $P'(x)$ is divisible by $(x-2)$ and $(x+2)$? The first line of math from this answer comes from the given information. If dividing by $(x-2)^3$ gives a remainder of $-2$, then $[P(x)+2]$ is divisible by $(x-2)^3$. Adriano then uses the product rule to show $P'(x)$ is then divisible by $(x-2)^2$. And similarly for the other factor. – 2017-02-23
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Hint: Take derivative of that polynomial to get $16 - 8 x^2 + x^4=(x^2-4)^2$.

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    Would the downvoter care to explain. Honestly, I have no clue. – 2017-02-23
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    Wasn't that polynomial arrived at by integrating $k(x-2)^2(x+2)^2$ in the first place? This appears to be circular logic. – 2017-02-23
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    @Mike You are probably right. I misread the (confusing because of the figures) question in the first place. I guess I deserve the downvote then,. – 2017-02-23